Minimum value of logarithmic equation

[ritwik06]

Homework Statement

For x>1, find the minimum possible value of [tex]2 log_{2}x-log_{x}(0.01)[/tex]

The Attempt at a Solution

Greater the value of x, greater is the value of expression. Right?
I tried to differentiate it, but it was no help. The derivative becomes zero when |log x|=[tex]\sqrt{log 2}[/tex]

Help me further.

 

[Defennder]

What did you get when you differentiate it?

 

[ritwik06]

What did you get when you differentiate it?

Why do you ask that? Is it wrong?
I got:
2*(1/x ln 2)-2*(1/log^{2} x)*(1/(x ln 10))

 

[Defennder]

Well, apparently I can't tell if it's correct unless I know your working. I can't read what you wrote that. Is it [tex]\frac{2 \ln x}{\ln 2} - \frac{2}{\log_2 x} \left( \frac{1}{x \ln 10} \right) [/tex].

If so then I don't think it's correct.

 

[ritwik06]

Well, apparently I can't tell if it's correct unless I know your working. I can't read what you wrote that. Is it [tex]\frac{2 \ln x}{\ln 2} - \frac{2}{\log_2 x} \left( \frac{1}{x \ln 10} \right) [/tex].

If so then I don't think it's correct.

[tex]\frac{2}{x ln 2}-\frac{2}{(log^{2} x)*(x ln 10)}[/tex]

 

[dynamicsolo]

This problem sets a similar trap to that of another problem you asked about. It would make life easier if you rewrote the [tex]log_{x}(0.01)[/tex] term as a log-base-2 term first, so you could combine it with the first term...

(You could grind through the differentiation you have, but it is a rather cumbersome "hammer" to use on the problem.)

 

[happyg1]

I got:
[tex]\frac{2}{x ln 2}-\frac{2 ln 10}{x ln^2 x}[/tex]
for the derivative.
I set it equal to zero, cross multiplied and came up with:

[tex]ln^2x=ln 10 ln 2[/tex]

Dunno if that helps...your derivative was different than mine.
CC

 

[ritwik06]

rewriting:
[tex]2log_{2}x+\frac{2(1+log_{2}5)}{log_{2}x}[/tex]

 

[happyg1]

Take the square root of both sides, ignore the absolute value bars, because for x>1 the thing is positive, then take the exponential of both sides. I got x=3.53722...
plug that back into get the y value.
CC